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Mark Kac wrote a paper about asymptotics of determinants whose main diagonal is taken from a function $f$, with $-1$ on the super and sub-diagonals. Specifically, $$ D_n = \begin{vmatrix} f(1/n) & -1 & 0 & \cdots & &0 \\ -1 & f(2/n) & -1 & \cdots \\ 0& -1 & f(3/n) \\ &&& \ddots \\ &&&& f((n-1)/n) & -1 \\ &&&& -1 & f(1) \end{vmatrix} $$ He then writes, "We begin with the elementary formula $$\frac{1}{\sqrt{D_n}} = \frac{1}{(\sqrt{\pi})^n} \int_{-\infty}^{\infty} \cdots \int_{-\infty}^{\infty} \exp\left[-\sum_{k=1}^n -\frac{1}{2}f\left(\frac{k}{n}\right) x_k^2 + 2\sum_{k=1}^{n-1} x_kx_{k+1}\right]dx_1dx_2\cdots dx_n $$

My question is, Where does this elementary formula come from??

The paper is: Asymptotic behavior of a class of determinants, L'Enseignement Mathematique, pp 177-183, 15(1969)

Tsintsadze
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Tyler McMillen
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  • You have a symmetric matrix there (a quadratic form) should be unitarily diagonalizable, that should suggest the suitable change of variable. – r9m Dec 04 '15 at 20:59
  • I'm not entirely sure where the $\sqrt{n}$ comes from (could you re-check that please?), but the form of the formula suggests the normalisation condition for a gaussian random vector with mean $0$ and covariance $M_n^{-1}$, where $M_n$ is the matrix above ($D_n = \det(M_n)$). – stochasticboy321 Dec 04 '15 at 21:18
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    Details can be found here: http://math.stackexchange.com/questions/126227/reference-for-multidimensional-gaussian-integral?rq=1 – r9m Dec 04 '15 at 21:26

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